Self-similarity of the diffusion equation

Question

I am going through this book Simulation of Complex Systems. In the chapter on Brownian Dynamics, we considered a "free diffusion" given by the Stochastic differential equation: $$\dot{x}(t)=W(t)$$ where $W(t)$ is a Gaussian white noise, with zero mean and unit variance.

For the purpose of simulation, we discretize the above as: $$x_{i+1}=x_i+W_i.\Delta t$$ with $\langle x_n\rangle=0$.

Now we can show that the Mean Squared Displacement (MSD) is proportional to $\Delta t$ !

Since physical reality should not depend on simulation parameter, we re-scale $W_i$ as: $$W_i\rightarrow \frac{w_i}{\sqrt{\Delta t}}$$ ($w_i$ having same property as $W_i$).

Finally our actual difference equation for diffusion with white noise: $$ x(t+\Delta t)=x(t) + w_i\sqrt{\Delta t}\tag{1} $$ Now, as mentioned in the beginning, here we are talking about free diffusion, so there is no characteristic time-scale in our system.

My question is, how can we show that our equation (1) is self-similar? That is, under a re-scaling of time the equation does not changes.

My understanding:

If we take $t\rightarrow\nu t$, so $\Delta t \rightarrow \nu\Delta t$, our equation (1) becomes $$ x(t+\Delta t)=x(t) + w_i\sqrt{\nu}\sqrt{\Delta t} $$ but $w_i$ is just the noise. And $\sqrt{\nu}w_i$ is also a noise with changed strength. So we may as well rename the noise as $w_i'$.

Is it valid?

Also, if $t\rightarrow\nu t$, what about $x(\nu t)$ and $x(\nu t+\nu\Delta t)$? The particle has shifted. The situation should not be same!

Answer 1

You are describing the Wiener process. Your notation is a bit confusing, since $W$ is usually what you call $x$. Yes, your system is self similar, but if you rescale time, you also need to rescale space: $$ \begin{align} \Delta t&\to\lambda \Delta t & \Delta x\to\sqrt \lambda \Delta x \end{align} $$ This comes from the fact that you need to conserve the diffusion coefficient, whose dimension is $[D] = L^2/T$. You can see this in your discretised equation: $$ x_{n+1} = x_n+ D\eta_n\Delta t $$ with $(\eta_n)_n$ iid gaussians with covariance: $$ \langle \eta_m\eta_n\rangle = \frac1{\Delta t}\delta_{mn} $$ It's easy to check that the scaling preserves the increment's equation.

Technically, the previously described scaling is a bit trivial and comes from the redundancy fo the parameters. The real self-similarity in the discrete setting is rather (taking $x_0=0$) that for $\lambda\in\mathbb N^*$, $(x_n)_n$ and $(\sqrt\lambda x_{\lambda n})_n$ describe the same process. This is why in the continuum limit, you'd expect to recover the full scaling invariance for any $\lambda\in\mathbb R_+^*$ for $x(t)$ and $\sqrt\lambda x(\lambda t)$. The magic of the random walk comes from the fact that for an arbitrary discrete noise, you don't have the discrete scaling invariance (the noise's distribution will not be stable under addition unlike the gaussian), but in the continuous limit, you do recover the scale invariance.

This is somewhat surprising coming from calculus. Indeed, if $x$ is differentiable, you'd expect the increments $\Delta t$ and $\Delta x$ to have the same scaling exponent, or at least that the scaling exponent of $x$ is a positive integer multiple of the one of $t$. On the contrary, the latter is half of the former. This rather shows that the trajectories are typically "rough"; quantitatively, they have a Holder exponent of $1/2$.